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Math Help - Cyclic groups

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    Cyclic groups

    Let G=Sym(n) .Let C be a cyclic group. Suppose that \theta \in Hom(C, G). then how to show that the generators for \theta (C) in G all have the same cycle type and so are conjugate in G.
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    Quote Originally Posted by dimuk View Post
    Let G=Sym(n) .Let C be a cyclic group. Suppose that \theta \in Hom(C, G). then how to show that the generators for \theta (C) in G all have the same cycle type and so are conjugate in G.
    I am assuming that the cyclic group is finite.

    If a generates \theta (C) then a^j generates \theta (C) where j is relatively prime with the order of a because a homomorphic image of a cyclic group is cyclic. Let a=\sigma_1\sigma_2 ... \sigma_k for disjoint cycles \sigma_i \in S_n. The order of a is the least common multiple of the lengths of all these cycles. Therefore, |\sigma_i| divides |a| for all i. Now if j is an integer so that (j,|a|) = 1 then certainly (j,|\sigma_i|)=1. Therefore, if \sigma_i is an i-cycle then \sigma_i^j still stays as an i-cycle. Thus, we have a^j = \sigma_1^j ... \sigma_k^j where \sigma_i is same cycle length as \sigma_i^j which means a and a^j have the same cycle type and consequently they are conjugate.
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